1. what does e^x mean?

for something like this: 12e^x-20=0 how do i find x?

2. i guess i'm answering my own question but i'm assuming i take the ln of both sides so essentially it would be ln (40/12) which is 1.2039?

3. Originally Posted by frozenflames
for something like this: 12e^x-20=0 how do i find x?
Do you realize that you posted this question in the Pre-Calculus forum?
Are you doing a course in Pre-Calculus?
If so, surely you know what $e^x$ means.

4. Originally Posted by frozenflames
i guess i'm answering my own question but i'm assuming i take the ln of both sides so essentially it would be ln (40/12) which is 1.2039?
This is correct, but like Plato said your thread title is strange. $e$ is just a number, so $e^x$ is $(e*e*e*e...)$, x times.

5. Originally Posted by frozenflames
for something like this: 12e^x-20=0 how do i find x?
Originally Posted by frozenflames
i guess i'm answering my own question but i'm assuming i take the ln of both sides so essentially it would be ln (40/12) which is 1.2039?

6. Originally Posted by mr fantastic
Spoiler:
$12e^x-20=0$

$12e^x=20 \Rightarrow e^x=\frac{20}{12}$

$\ln(e^x)=\ln(\frac{20}{12}) \Rightarrow x\ln(e)= \ln(\frac{20}{12})$

I get the same answer as OP. What did I do wrong?

7. Originally Posted by Jameson
Spoiler:
$12e^x-20=0$

$12e^x=20 \Rightarrow e^x=\frac{20}{12}$

$\ln(e^x)=\ln(\frac{20}{12}) \Rightarrow x\ln(e)= \ln(\frac{20}{12})$

I get the same answer as OP. What did I do wrong?
What you got (and which is correct) is
Spoiler:
$x = \log_e \frac{20}{12}$

This is not the same as $\log_e \frac{40}{12}$ (which is what the OP got) ....!